13th IPhO 1983 Theo II Electricity Solution

IPhO 1983 Theoretical Question II

Electricity – Problem II - Solution Page 1 from 12

2.Electricity – Problem II (8 points) Different kind of oscillation Let’s consider the electric circuit in the figure, for which mHL 101 = ,

mHL 202 = , nFC 101 = , nFC 52 = and Ω= kR 100 . The switch K being closed the circuit is coupled with a source of alternating current. The current furnished by the source has constant intensity while the frequency of the current may be varied.

a. Find the ratio of frequency mf for which the active power in circuit has the maximum value mP and the frequency difference

−+ −=∆ fff of the frequencies +f and −f for which the active power in the circuit is half of the maximum power mP .

The switch K is now open. In the moment 0t immediately after the switch is open the intensities of the currents in the coils 1L and

Ai 1,001 = and Ai 2,002 = 1L (the currents flow as in the figure); at the same moment, the potential difference on the capacitor with capacity 1C is Vu 400 = :

b. Calculate the frequency of electromagnetic oscillation in 2211 LCCL circuit;

c. Determine the intensity of the electric current in the AB conductor;

d. Calculate the amplitude of the oscillation of the intensity of electric current in the coil 1L .

Neglect the mutual induction of the coils, and the electric resistance of the conductors. Neglect the fast transition phenomena occurring when the switch is closed or opened. Problem II - Solution a. As is very well known in the study of AC circuits using the formalism of complex numbers, a complex inductive reactance jLX L ⋅⋅= ω , ( 1−=j ) is attached to the inductance L - part of a circuit supplied with an alternative current having the pulsation ω .

Similar, a complex capacitive reactance ω⋅

−=C

jX C is attached to the capacityC .

A parallel circuit will be characterized by his complex admittanceY . The admittance of the AC circuit represented in the figure is



( )

+−+⋅+=

⋅−

⋅−

⋅⋅+

⋅⋅+=

2121

21

21

111

111

LLCCj

RY

jC

jC

jLjLRY ωω

ωω ( 2.1)

The circuit behave as if has a parallel equivalent capacity C

21 CCC += ( 2.2)

and a parallel equivalent inductance L

+=

+=

21

21

21

111

LLLL

L

LLL ( 2.3)

The complex admittance of the circuit may be written as

⋅−⋅⋅+=

ωω

LCj

RY 11 ( 2.4)

and the complex impedance of the circuit will be

⋅−⋅+

⋅−

⋅⋅+

=

=

22 11

11

1

ωω

ωω

LC

R

CL

jRZ

YZ

( 2.5)

The impedance Z of the circuit, the inverse of the admittance of the circuit Y is the modulus of the complex impedance Z

Y

LC

R

ZZ 1

11

122=

⋅−⋅+

==

ωω

( 2.6)

The constant current source supplying the circuit furnish a current having a momentary value ( )ti

( ) ( )tIti ⋅⋅⋅= ωsin2 , ( 2.7)

where I is the effective intensity (constant), of the current and ω is the current pulsation (that can vary) . The potential difference at the jacks of the circuit has the momentary value ( )tu

( ) ( )ϕω +⋅⋅⋅= tUtu sin2 ( 2.8)

where U is the effective value of the tension and ϕ is the phase difference between tension and current. The effective values of the current and tension obey the relation



ZIU ⋅= ( 2.9) The active power in the circuit is

RIZ

RUP

222 ⋅== ( 2.10)

Because as in the enounce,

==

constantRconstantI

( 2.11)

the maximal active power is realized for the maximum value of the impedance that is the minimal value of the admittance . The admittance

22 11

⋅−⋅+

=

ωω

LC

RY ( 2.12)

has– as function of the pulsation ω - an „the smallest value”

RY 1

min = ( 2.13)

for the pulsation

CLm ⋅=

1ω ( 2.14)

In this case

01=

⋅−⋅

ωω

LC . ( 2.15)

So, the minimal active power in the circuit has the value 2IRPm ⋅= ( 2.16)

and occurs in the situation of alternative current furnished by the source at the frequency mf

LCf mm ⋅⋅

==π

ωπ 2

121 ( 2.17)

To ensure that the active power is half of the maximum power it is necessary that

==

⋅=⋅

=

222

222

1221

21

YZR

IRR

IZ

PP m

( 2.18)

That is



⋅−⋅=±

⋅−⋅+=

ωω

ωω

LC

R

LC

RR11

112 2

22 ( 2.19)

The pulsation of the current ensuring an active power at half of the maximum power must satisfy one of the equations

0112 =⋅

−⋅

±CLCR

ωω ( 2.20)

The two second degree equation may furnish the four solutions

CLCRCR ⋅+

⋅±

⋅±=

4121

21 2

ω ( 2.21)

Because the pulsation is every time positive, and because

CRCLCR ⋅>

⋅+

⋅141 2

( 2.22)

the only two valid solutions are

CRCLCR ⋅±

⋅+

⋅=± 2

14121 2

ω ( 2.23)

It exist two frequencies ±± = ωπ21f allowing to obtain in the circuit an active power representing half of

the maximum power.

⋅−

⋅+

⋅=

⋅+

⋅+

⋅=

−

+

CRCLCRf

CRCLCRf

2141

21

21

2141

21

21

2

2

π

π ( 2.24)

The difference of these frequencies is

CRfff

⋅=−=∆ −+

121π

( 2.25)

the bandwidth of the circuit – the frequency interval around the resonance frequency having at the ends a signal representing 21 from the resonance signal. At the ends of the bandwidth the active power reduces at the half of his value at the resonance. The asked ratio is



( ) ( )

⋅+⋅+

=∆

=⋅⋅

=∆

21

2121

LLLLCCR

ff

LCR

CLCR

ff

m

m

( 2.26)

Because

*

=

=

mHL

nFC

32015

it results that 1510 −⋅= sradmω

and

15010201015310100 3

93 =

××⋅

⋅×==∆ −

−

LCR

ffm ( 2.27)

The (2.26) relation is the answer at the question a. b. The fact that immediately after the source is detached it is a current in the coils, allow as to admit that currents dependents on time will continue to flow through the coils.

Figure 2.1

The capacitors will be charged with charges variable in time. The variation of the charges of the capacitors will results in currents flowing through the conductors linking the capacitors in the circuit. The momentary tension on the jacks of the coils and capacitors – identical for all elements in circuit – is also dependent on time. Let’s admit that the electrical potential of the points C and D is )(tu and the



potential of the points A and B is zero. If through the inductance 1L passes the variable current having the momentary value ( )ti1 , the relation between the current and potentials is

( ) 011 =−

dtdiLtu ( 2.28)

The current passing through the second inductance ( )ti2 has the expression,

( ) 022 =−

dtdiLtu ( 2.29)

If on the positive plate of the capacitor having the capacity 1C is stocked the charge ( )tq1 , then at the jacks of the capacitor the electrical tension is ( )tu and

uCq ⋅= 11 ( 2.30)

Deriving this relation it results

dtduC

dtdq

⋅= 11 ( 2.31)

But

31 i

dtdq

−= ( 2.32)

because the electrical current appears because of the diminishing of the electrical charge on capacitor plate. Consequently

dtduCi ⋅−= 13 ( 2.33)

Analogous, for the other capacitor,

dtduCi ⋅−= 44 ( 2.34)

Considering all obtained results

=

=

2

2

1

1

Lu

dtdi

Lu

dtdi

( 2.35)

respectively

=

−=

2

2

24

2

2

13

dtudC

dtdi

dtudC

dtdi

( 2.36)

Denoting ( )ti5 the momentary intensity of the current flowing from point B to the point A , then the same momentary intensity has the current through the points C and D . For the point A the Kirchhoff rule of the currents gives



351 iii =+ ( 2.37)

For B point the same rule produces

254 iii =+ ( 2.38)

Considering (2.37) and (2.38) results

2431 iiii −=− ( 2.39)

and deriving

dtdi

dtdi

dtdi

dtdi 2431 −=− ( 2.40)

that is

( )

+⋅=

+⋅−

+=−−

212

2

21

2

2

22

2

121

11 CCdt

udLL

u

dtudC

dtudC

Lu

Lu

( 2.41)

Using the symbols defined above

=+

⋅=−

012

2

uLC

u

Cdt

udLu

( 2.42)

Because the tension obeys the relation above, it must have a harmonic dependence on time ( ) ( )δω +⋅⋅= tAtu sin ( 2.43)

The pulsation of the tension is

CL ⋅=

1ω ( 2.44)

Taking into account the relations (2.43) and (2.36) it results that

( )( ) ( )

( )( ) ( )

+⋅⋅⋅⋅−=+⋅⋅−=

+⋅⋅⋅⋅−=+⋅⋅−=

δωωδω

δωωδω

tACtAdtdCi

tACtAdtdCi

cossin

cossin

224

113

( 2.45)

and

( )

( )

+⋅⋅⋅==

+⋅⋅⋅==

δω

δω

tALL

udtdi

tALL

udtdi

sin1

sin1

22

2

11

1

( 2.46)

It results that



( )

( )

++⋅⋅⋅⋅

=

++⋅⋅⋅⋅

=

NtAL

i

MtAL

i

δωω

δωω

cos1

cos1

22

11

( 2.47)

In the expression above, A , M , N and δ are constants that must be determined using initially conditions. It is remarkable that the currents through capacitors are sinusoidal but the currents through the coils are the sum of sinusoidal and constant currents. In the first moment

( )( )( )

======

AiiAiiVuu

2,001,00

400

022

011

0

( 2.48)

Because the values of the inductances and capacities are

====

nFCnFC

HLHL

510

02,001,0

2

1

2

1

( 2.49)

the equivalent inductance and capacity is

=××

=

+⋅

=

+=

−

−

HHL

LLLLL

LLL

1501

103102

111

2

4

21

21

21

( 2.50)

respectively

=+=nFC

CCC15

21 . ( 2.51)

From (2.44) results

15

9

101015

1501

1 −

−

⋅=×⋅

= sradω ( 2.52)

The value of the pulsation allows calculating the value of the requested frequency b. This frequency has the value

*

f

Hzfππ

ω210

2

5

== ( 2.53)

*

c. If the momentary tension on circuit is like in (2.43), one may write



( ) ( )

( )

=

=⋅=

Au

uAu

0

0

sin

sin0

δ

δ ( 2.54)

From the currents (2.47) is possible to write

( )

( )

+⋅⋅⋅

=

+⋅⋅⋅

=

NAL

i

MAL

i

δω

δω

cos1

cos1

202

101

( 2.55)

On the other side is possible to express (2.39) as

( ) ( )

( ) ( )

−+⋅⋅⋅⋅

−+⋅⋅⋅⋅−

=+⋅⋅⋅⋅+++⋅⋅⋅⋅

−=−

NtAL

tAC

tACMtAL

iiii

δωω

δωω

δωωδωω

cos1cos

coscos1

22

11

2431

( 2.56)

An identity as DCBA +⋅≡+⋅ αα coscos ( 2.57)

is valuable for any value of the argument α only if

==

DBCA

( 2.58)

Considering (2.58), from (2.56) it results

( )

+⋅−=+⋅⋅

=+

2121

110

LLACCA

NM

ωω

( 2.59)

For the last equation it results that the circuit oscillate with the pulsation in the relation (2.44) Adding relations (2.55) and considering (2.54) and (2.59) results that

( )

( )

( )

⋅⋅+=

+⋅⋅

+=

+⋅⋅

+=

+⋅⋅⋅=+

ALii

LLA

iiLL

iiA

LLAii

ωδ

ω

δ

ωδ

ωδ

0201

11

0201

11

0201

110201

cos

111cos

111cos

111cos

( 2.60)



The numerical value of the amplitude of the electrical tension results by summing the last relations from (2.54) and (2.60)

( )

( )

( )( ) ( )( )( )

( ) ( )( )

⋅⋅++=

=

⋅⋅+

+

=+

⋅⋅+=

=

20201

20

2

0201

2

0

22

0201

0

1

1sincos

cos

sin

ω

ω

δδ

ωδ

δ

LiiuA

ALii

Au

ALii

Au

( 2.61)

The numerical value of the electrical tension on the jacks of the circuit is

( ) ( )

( ) ( )

=+=

⋅⋅+=

VA

A

264020040

10150

13,040

22

252

( 2.62)

And consequently from (2.54) results

( )

( )

==

=

261

264040sin

sin 0

δ

δAu

( 2.63)

and

( )265cos =δ ( 2.64)

Also

( )

( )

=

=

5151

arctg

tg

δ

δ ( 2.65)

From (2.55)

( )

( )

⋅⋅⋅

−=

⋅⋅⋅

−=

δω

δω

cos1

cos1

202

101

AL

iN

AL

iM ( 2.66)

the corresponding numerical values are



=

⋅⋅

⋅−=

−=

⋅⋅

⋅−=

AAN

AAM

1,02652640

1002,012,0

1,02652640

1001,011,0

5

5

( 2.67)

The relations (2.47) becomes

*

( )( )

( )( )

+=

++⋅⋅=

−=

−+⋅⋅=

025

2

015

1

~1,05110cos100

262

~1,05110cos100

264

IiAarctgti

IiAarctgti ( 2.68)

The currents through the coils are the superposition of sinusoidal currents having different amplitudes and a direct current passing only through the coils. This direct current has the constant value

AI 1,00 = ( 2.69)

as in the figure 2.2.

*

Figure 2.2

The alternative currents through the coils has the expressions

( )( )

( )( )

+⋅⋅=

+⋅⋅=

Aarctgti

Aarctgti

5110cos100

262~

5110cos100

264~

52

51

( 2.70)



The currents through the capacitors has the forms

( )( )( )( )( )

( )( )( )( )( )

+⋅−=

+⋅⋅⋅×−=

+⋅−=

+⋅⋅⋅×−=

−

−

Aarctgti

Aarctgti

Aarctgti

Aarctgti

5110cos100

262

5110cos2640105

5110cos100

264

5110cos26401010

54

544

53

543

( 2.71)

The current 5i has the expression

( )( )

++⋅−=

−=

Aarctgti

iii

1,05110cos100

268 55

135

( 2.72)

The value of the intensity of 5i current is the answer from the question c. The initial value of this current is

AAi 3,01,0265

100268

5 −=

+−= ( 2.73)

d. The amplitude of the current through the inductance

*

1L is

( ) ( )( ) AAAarctgti 2,0100

2645110cos100

264max~max 51 ≈=

+⋅⋅= ( 2.74)

representing the answer at the question d.

*

Professor Delia DAVIDESCU, National Department of Evaluation and Examination–Ministry of Education and Research- Bucharest, Romania Professor Adrian S.DAFINEI,PhD, Faculty of Physics – University of Bucharest, Romania

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13th IPhO 1983 Theo II Electricity Solution